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  4. A capacitor of capacitance value 1 μ F is charged to 30 V and the battery is then disconnected. If the remaining circuit is connected across a 2 μ F capacitor, the energy lost by the system is

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Q. A capacitor of capacitance value $ 1\,\mu F $ is charged to $ 30 \,V $ and the battery is then disconnected. If the remaining circuit is connected across a $ 2\,\mu F $ capacitor, the energy lost by the system is

KEAMKEAM 2008Electrostatic Potential and Capacitance

Solution:

Energy $ {{E}_{1}}=\frac{1}{2}{{C}_{1}}V_{1}^{2}=\frac{1}{2}\times 1\times {{10}^{-6}}\times {{(30)}^{2}} $
$ =450\times {{10}^{-6}}J $ Common potential
$ V=\frac{{{q}_{1}}+{{q}_{2}}}{{{C}_{1}}+{{C}_{2}}} $
$ =\frac{1\times 30+0}{1+2}=10\,volt $
$ {{E}_{2}}=\frac{1}{2}({{C}_{1}}+{{C}_{2}}){{V}^{2}} $
$ =\frac{1}{2}(1+2)\times {{10}^{-6}}\times {{(10)}^{2}} $
$ =1.5\times 100\times {{10}^{-6}} $
$ =150\times {{10}^{-6}}J $
Loss of energy $ ={{E}_{2}}-{{E}_{1}}=300\,\mu J $